Problem 9, section 4.2 of R.M. Dudley, Real Analysis and Probability

Question

Let $f$ be a measurable function from $X$ onto $S$ where $(X,\mathcal A)$ is a measurable space and $(S, e)$ is a metric space with Borel $\sigma$-algebra. Let $T$ be a subset of $S$ with discrete relative topology (all subsets of $T$ are open in $T$ ). Show that there is a measurable function $g$ from $X$ onto $T$ . Hint: For $f(x)$ close enough to $t \in T$ , let $g(x) = t$; otherwise, let $g(x) = t_0$ for a fixed $t_0 \in T$ .

I made the following observations:

1. If $B\subset T$, then $B=U\cap T$ for some open set $U$ in $S$, and so $f^{-1}(B)=f^{-1}(U\cap T)=f^{-1}(U)\cap f^{-1}(T)\in\mathcal{A}$ if $f^{-1}(T)\in\mathcal{A}$. Hence $f^{-1}(T)\in\mathcal{A}$ implies that the preimage of any subset of $T$ under $f$ is measurable.

2. For $t_0\in T$ fixed, we can let $g(x)= f(x)$ if $x\in f^{-1}(T)$ and $g(x)=t_0$ otherwise. Then $g$ is onto $T$, and for $t\in T$ we have $g^{-1}(\{t\})=f^{-1}(\{t\})$ or $f^{-1}(\{t\})\cup (f^{-1}(T))^c$. It follows that $g^{-1}(\{t\})$ is measurable for all $t\in T$, and if $T$ is countable this is sufficient for the measurability of $g$.

But how to deal with the case where $T$ is uncountable?

Any help on this is very appreciated.

Answer 1

For each $t\in T$, choose an open set $U_t$ around $t$ so that $U_t \cap T = \{t\}$.
The hint is telling you to define $g(x) = t$ on the set $f^{-1}(U_t)$.

The technical difficulty is to ensure the sets $\{U_t : t\in T\}$ can be chosen mutually disjoint.

Answer 2

Since the relative topology of $T$ is discrete, we can choose, for each $t\in T$, an open set $U_t$ in $S$ such that

$$\{t\}=U_t\cap T$$

and since $S$ is a metric space we may assume without loss of generality that $U_t=B(t,r_t)$ for some $r_t>0$.

Let $s,t\in T$ with $s\neq t$. We claim that $B(t,r_t/2)\cap B(s,r_s/2)=\emptyset$. To see this, note that $s\notin B(t,r_t)$ and $t\notin B(s,r_s)$. Therefore $d(s,t)\geq\max(r_s,r_t)$. If $y\in B(t,r_t/2)\cap B(s,r_s/2)$, then from the triangle inequality we obtain $d(t,s)<r_t/2+r_s/2\leq \max(r_s,r_t)$, a contradiction. Hence we can assume without loss of generality that the $U_t$ are disjoints.

Now fix some $t_0\in T$ and define $g:X\to T$ by $g(x)=t$ if $x\in f^{-1}(U_t)$ for some $t\in T$ and $g(x)=t_0$ otherwise. This is well-defined since the $U_t$ are disjoints. The map $g$ is onto since $f$ is onto and $f(x)=t$ implies $x\in f^{-1}(U_t)$. Finally $g$ is measurable since for $B\subset T$ we have

$$g^{-1}(B)=\cup_{t\in B}f^{-1}(U_t)=f^{-1}(\cup_{t\in B} U_t)\in \mathcal A$$ if $t_0\notin B$ and

$$g^{-1}(B)=\cup_{t\in B}f^{-1}(U_t)\cup (X\setminus \cup_{t\in T}f^{-1}(U_t))$$

$$=f^{-1}(\cup_{t\in B} U_t) \cup (X\setminus f^{-1}(\cup_{t\in T}U_t))\in \mathcal A$$ if $t_0\in B$.